The working area allocation routines use 4 byte word alignment.
In the corner case the size of the working area is not aligned,
target_alloc_working_area() of size = target_get_working_area_avail()
will fail because the size gets aligned up and does not fit to the area
which size is aligned down.
Align down the result of target_get_working_area_avail() to cope with that
corner case.
While on it use fancy ALIGN_... macros instead of bitwise and operator.
Change-Id: Ia2a1e861c401c2c78fe6323379a3776fb4f47b06
Signed-off-by: Tomas Vanek <vanekt@fbl.cz>
Reviewed-on: https://review.openocd.org/c/openocd/+/7096
Tested-by: jenkins
Reviewed-by: Erhan Kurubas <erhan.kurubas@espressif.com>
Reviewed-by: Antonio Borneo <borneo.antonio@gmail.com>
struct working_area *new_wa = malloc(sizeof(*new_wa));
if (new_wa) {
new_wa->next = NULL;
- new_wa->size = target->working_area_size & ~3UL; /* 4-byte align */
+ new_wa->size = ALIGN_DOWN(target->working_area_size, 4); /* 4-byte align */
new_wa->address = target->working_area;
new_wa->backup = NULL;
new_wa->user = NULL;
}
/* only allocate multiples of 4 byte */
- if (size % 4)
- size = (size + 3) & (~3UL);
+ size = ALIGN_UP(size, 4);
struct working_area *c = target->working_areas;
uint32_t max_size = 0;
if (!c)
- return target->working_area_size;
+ return ALIGN_DOWN(target->working_area_size, 4);
while (c) {
if (c->free && max_size < c->size)
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